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baherus [9]
2 years ago
14

What's the chemical formula for salt, like is it NaCl?? the salt we eat???????​

Chemistry
2 answers:
GenaCL600 [577]2 years ago
7 0

Answer:

Yes

Explanation:

NaCl is table salt

Also known as sodium chloride

MA_775_DIABLO [31]2 years ago
3 0

yeah it's NaCl, I just looked it up

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How many grams of carbon should be burned in an excess of oxygen at stp to obtain 2. 21 l of carbon dioxide?.
snow_tiger [21]
You must burn 1.17 g C to obtain 2.21 L CO2 at
STP.
The balanced chemical equation is
C+02+ CO2.
Step 1. Convert litres of CO, to moles of CO2.
STP is 0 °C and 1 bar. At STP the volume of 1 mol
of an ideal gas is 22.71 L.
Moles of CO2= 2.21 L CO2 × (1 mol CO2/22.71 L
CO2) = 0.097 31 mol CO2
Step 2. Use the molar ratio of C:CO2 to convert
moles of CO to moles of C
Moles of C= 0.097 31mol CO2 × (1 mol C/1 mol
CO2) = 0.097 31mol C
Step 3. Use the molar mass of C to calculate the
mass of C
Mass of C= 0.097 31mol C × (12.01 g C/1 mol C) =
1.17 g C
It looks as if you are using the old (pre-1982)
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4 0
2 years ago
What number of moles of Pb (NO3)2 is consumed when 54 g of KNO3 is produced
Ivenika [448]

Explanation:

hope it helps you understand moles

5 0
3 years ago
Which of the following is a characteristic of alcohols
liq [111]
You forgot to add pictures ..
3 0
3 years ago
One mole of an ideal gas with a volume of 1.0 L and a pressure of 5.0 atm is allowed to expand isothermally into an evacuated bu
Deffense [45]

Answer:

w= - 1.7173 kJ, q= 1.7173 kJ, q(rev) = 1717.3 J = 1.7173 kJ.

Explanation:

Okay, from the question we are given the information below;

Number of moles, n= 1 mole; initial volume, v(1) = 1.0 litres (L); pressure (p) = 5atm, final volume(v2) = 2.0 Litres(L) ; the workdone, w= not given; the heat, q and q(rev)= not given and the gas was said to expand isothermally.

So, this question is a question from the part of chemistry known as thermodynamics. Therefore, grip yourself we are delving into thermodynamics 'waters' now.

For expansion isothermally; the workdone, w= -nRT ln v2/v1.

Where T= temperature= 25° C = 298 k and R= gas constant.

Therefore; workdone, w = - 1 × 8.314 × 298 × ln(2/1).

Workdone,w= - 1717.32204643. =

- 1717.3 Joules (J).

==> Workdone,w= - 1.7173 kJ.

Then, we are to find q. q can be solved by using the first law of thermodynamics, which by mathematical representation is:

∆U= q + w. Where ∆U= change in internal enegy. Since the question is dealing with isothermal expansion, there is this rule that says for an isothermal expansion ∆U = 0.

Hence, 0 =q + [- 1717.3 Joules (J)].

q=1717.3 J = 1.7173 kJ.

Finally, the q(rev) which is= nRT ln (v2/V1).

q(rev) = 1 × 8.314 × 298 ln (2/1).

q(rev) = 1717.3 J = 1.7173 kJ.

PS: please note the negative signs in the workdone and the positive sign in the q(rev).

7 0
3 years ago
as the elements period 3 are considered in order of increasing atomic number, the number of principal energy levels in each succ
Diano4ka-milaya [45]

Answer:

stay the same.

Explanation:  Period 3 consists of the full 1s, 2s, and 2p electron orbitals, plus the 3s and 3p valence orbitals, which are filled with a total of 8 more electrons as we move from left (Na) to the far right (Ar):

Na:  1s2 2s2 2p6 3s1

Ar:    s2 2s2 2p6 3s2 3p6

As we move from left to right, and ignoring the already-filled 1s, 2s, and 2p orbitals, the period three starting and ending elements have the following:

Na: 3s1

Ar: 3s2, 3p6

All the new electrons electrons filled the third energy level (3s and 3p).  So the energy level does not change, just the orbitals.

5 0
3 years ago
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